About the Fourier transform of the sign function I'm trying to calculate the Fourier transform of the function $f(x):=sign(x)$.
I have read some texts where this is solved approximating the function $f$ by other functions, $f_a$, defined as follows $f_a(x)=e^{-a|x|}$.
Then, the say that $TF(f)=TF(\displaystyle\lim_{a\to 0}f_a)=\displaystyle\lim_{a\to 0}(TF(f_a))$. They easily compute $TF(f_a)$ and take the limit to conclude.
My questions are: 
$\bullet$ How can we justify the change of the limit with the $TF$?
$\bullet$ Is there another more rigurous way to solve this, maybe considering $f$ as a distribution instead of a function?
Thanks a lot for any help.
 A: Concerning the first question, the Fourier transform is continuous as a map $\mathscr{S}\to \mathscr{S}$ from the Schwartz space of rapidly decreasing functions to itself, hence its transpose, the Fourier transform of tempered distributions is also continuous. Now the functions $f_a$ converge locally uniformly and monotonically to the sign function, hence in the sense of tempered distributions, therefore moving the $\lim$ past $TF$ is legitimate. Also, the $f_a$ are all integrable functions, hence their Fourier transform in the sense of tempered distributions can be naturally identified with their $L^1$ Fourier transforms, and it remains to identify the limit of $TF(f_a)$ in the sense of tempered distributions.
Concerning the second question, the tempered distributions that we can represent as locally integrable functions (with appropriate growth conditions) are particularly nice, since you can do more explicit computations with these. 
The given argument does - at least when made precise - view the involved functions as tempered distributions, otherwise taking $\lim\limits_{a\to 0} TF(f_a)$ would be problematic.
One could also resort to the formal definition of the Fourier transform of a tempered distribution,
$$TF(f)[\varphi] = f[\hat{\varphi}]$$
for $\varphi\in \mathscr{S}$, and see whether computing with that allows one to find an explicit representation of the Fourier transform of the sign function. It does, but the approximation argument is simpler.
